4
$\begingroup$

In my calculations I have these combinations

PauliMatrix[1].f PauliMatrix[2]

where f is a scalar function. The result is as if Mathematica assumes f to be a matrix:

{{0, -I {{0, 1}, {1, 0}}.f}, {I {{0, 1}, {1, 0}}.f, 0}}

I tried to attach a unit matrix to f:

PauliMatrix[1]  .f PauliMatrix[0] .PauliMatrix[2]

The result is the same.

Does anyone see a way to get the same output as from

In[131]:= f PauliMatrix[1]. PauliMatrix[2]

Out[131]= {{I f, 0}, {0, -I f}}
$\endgroup$
1
  • 4
    $\begingroup$ Put parantheses around f PauliMatrix[2]'. It's interpreting the expression as dotting PauliMatrix[1]` with f and then multiplying element-wise with PauliMatrix[2], due to order-of-operations. $\endgroup$
    – march
    Jun 27, 2015 at 21:19

1 Answer 1

7
$\begingroup$

In Mathematica, the type of variable is interpreted based on the context, and if there are no values associated with the variable, then often nothing is done. When you write PauliMatrix[1].f, since there are no values/rules associated with f, this just returns

{{0, 1}, {1, 0}}.f

because the function Dot doesn't evaluate unless the arguments are vectors, matrices, or tensors (essentially, Lists, I think).

When you then write

PauliMatrix[1].f PauliMatrix[2]

it interprets {{0, 1}, {1, 0}}.f as a single object and multiplies this object as a scalar by {{0, -I}, {I, 0}}, yielding

{{0, -I {{0, 1}, {1, 0}}.f}, {I {{0, 1}, {1, 0}}.f, 0}}

The problem is order of operations, essentially: your expression Dots before it multiplies. Therefore, all you really need is a set of parantheses:

PauliMatrix[1].(f PauliMatrix[2])

which yields

{{I f, 0}, {0, -I f}}
$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.